advanced35 min studyLesson 519 in Mathematics

Symmetric-group representations and Young diagrams

Label complex symmetric-group irreducibles by partitions and compute dimensions using actual hook products.

On this page

What you’ll understand

  • Distinguish a partition shape from its tableaux.
  • State the complex Specht-module classification with its field restriction.
  • Compute hook-length dimensions and cross-check small regular decompositions.

The symmetric group has a uniform description of every complex irreducible representation using partitions and Young diagrams. A partition supplies the shape, while tableaux fill that shape with labels and organize the corresponding linear algebra. Row symmetrization and column antisymmetrization construct the relevant group-algebra subspace. The classification and dimension formula are substantial theorems, not consequences of drawing a diagram alone. Here we state their hypotheses, explain the construction, and carry out hook calculations that can be checked cell by cell.

Shapes label types, tableaux label arrangements

A partition of n is a decreasing list of positive row lengths summing to n. Its Young diagram has those left-justified rows. A tableau fills the boxes with the numbers one through n without repetition. A standard tableau has entries increasing across every row and down every column. Different fillings of a fixed shape do not label different complex irreducible types; they help construct or index vectors associated with that same type. The distinction between a shape and a filling is essential when counting dimensions.

Fix one filling and consider permutations preserving each row as a set, and permutations preserving each column as a set. Sum the row permutations without signs to get a row symmetrizer, and sum the column permutations with their signs to get a column antisymmetrizer. Their product is a Young symmetrizer, with its order declared because the factors generally do not commute. We use unnormalized sums and put the row factor on the left. The left ideal generated by that element is preserved by left multiplication by every permutation.

aT=∑r∈RTr,bT=∑c∈CTsgn⁡(c)c,cT=aTbT,Sλ=C[Sn]cTa_T=\sum_{r\in R_T}r,\qquad b_T=\sum_{c\in C_T}\operatorname{sgn}(c)c,\qquad c_T=a_Tb_T,\qquad S^\lambda=\mathbb C[S_n]c_T
The row and column sets are subgroups for a fixed tableau T of shape λ. All statements about irreducibility here are over the complex numbers.↔ Scroll wide equations horizontally.

The classification is a theorem over the complex field

The complex Specht-module classification says each left ideal just constructed is irreducible, every complex irreducible of the symmetric group arises in this way, and two shapes give equivalent modules exactly when their partitions agree. Changing the filling changes the presentation but preserves the type. We invoke this theorem rather than presenting its full straightening and irreducibility proof. The standard-basis theorem further identifies the dimension with the number of standard tableaux of the shape, providing a combinatorial meaning for a matrix dimension.

A single row gives the trivial representation because all labels are symmetrized, and a single column gives the sign representation because all labels are antisymmetrized. Other shapes mix these constraints in subtler ways. In characteristic dividing the group order, these same constructions require different irreducibility statements: Specht modules can be reducible, and partitions do not give the same simple one-to-one classification of irreducibles. The complex theorem cannot simply be transferred by reducing its matrices modulo a prime. Maschke's characteristic restriction explains part of that difficulty.

hij=1+(λi−j)+#{k>i:λk≥j},dim⁡CSλ=n!∏(i,j)∈λhijh_{ij}=1+(\lambda_i-j)+\#\{k>i:\lambda_k\ge j\},\qquad \dim_{\mathbb C}S^\lambda=\frac{n!}{\prod_{(i,j)\in\lambda}h_{ij}}
A hook counts the cell itself, the cells to its right, and the cells below it in the same column. The hook-length theorem counts standard tableaux and hence the complex irreducible dimension.↔ Scroll wide equations horizontally.

Hooks give a stringent arithmetic check

The hook length counts only the arm to the right and the leg below, not an entire southeast rectangle. Computing each box separately avoids multiplying incorrect row lengths or counting the corner twice. For a single row the hook product is the factorial, giving dimension one; a single column gives the same result. Transposing any shape preserves its hook multiset and therefore its dimension, though the corresponding representation need not be equivalent. In fact transposition corresponds to tensoring with the sign representation over the complex numbers.

The group order supplies another independent check. List all partitions of a small n, compute their hook dimensions, and sum the squares. Completeness of the complex classification requires that sum to equal the factorial. For larger n the hook formula is far more efficient than constructing all permutations, but it still does not provide every character value at other conjugacy classes. Diagram combinatorics controls several different quantities, and a dimension formula should not be mistaken for a full character formula or an explicit basis of matrices.

Compare two six-box shapes

Compute the dimensions of the complex S6S_6 irreducibles labeled by (4,2)(4,2) and (3,2,1)(3,2,1). For the first shape, explain why the lower-left hook has length two rather than three.

  1. For (4,2)(4,2), the top-row hooks are 5,4,2,15,4,2,1 and the bottom-row hooks are 2,12,1. The lower-left cell sees itself and the cell to its right, with nothing below; boxes above never enter its hook.

  2. The product is 5⋅4⋅2⋅1⋅2⋅1=805\cdot4\cdot2\cdot1\cdot2\cdot1=80, so the dimension is 6!/80=96!/80=9. For (3,2,1)(3,2,1), the top row gives hooks 5,3,15,3,1, the next row 3,13,1, and the last row 11.

  3. The second product is 4545, giving dimension 720/45=16720/45=16. The transpose (2,2,1,1)(2,2,1,1) of the first shape has the same dimension nine and represents its sign twist, not an automatic copy of the same irreducible.

The result

The two shapes have dimensions nine and sixteen. Counting only cells to the right or below produces integer hook quotients with a clear tableau interpretation.

Explore the model

Interactive lab

Interactive lab · waves

Symmetric and antisymmetric signs in the smallest case

Pause and Reset, then set both phase 00 degrees and amplitude ratio 11, predicting amplitude 22. Reset and set both phase 180180 degrees and ratio 11, predicting cancellation to amplitude 00. This is a two-component phase analogy for adding equal contributions versus their negatives in the row and column operators for S2S_2. The model does not construct Young symmetrizers or count tableaux; compute those algebraically. Changing a parameter retains the old clock, so Reset starts each comparison.

Symmetric and antisymmetric signs in the smallest caseTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRow addition and column signs organize different symmetry constraints; their higher-dimensional interaction needs algebra, not the phase analogy alone.

Check your understanding

What is the complex irreducible dimension for the S6 shape (4,2), whose hook lengths are 5,4,2,1,2,1?

Try it yourself

Compute the complex irreducible dimensions for all five partitions of four and verify their squared sum is ∣S4∣|S_4|. Show the hook product for (2,2)(2,2) explicitly.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The shapes are (4), (3,1), (2,2), (2,1,1), and (1,1,1,1); transposed shapes have equal dimensions.

Compare with the solution

The single row and column have dimension one. For (3,1)(3,1) the hooks are 4,2,1,14,2,1,1, product eight, giving dimension three; its transpose also has dimension three. For (2,2)(2,2) the hooks are 3,2,2,13,2,2,1, product twelve, giving dimension two. Thus the dimensions are 1,3,2,3,11,3,2,3,1, and 12+32+22+32+12=241^2+3^2+2^2+3^2+1^2=24. This verifies the regular dimension identity for the complete complex list.

Take it with you

  • Partitions label complex irreducible types; tableaux are fillings of those shapes.
  • Complex Specht irreducibility is a theorem with field-sensitive hypotheses.
  • Hook products count dimensions and can be checked against the regular sum of squares.

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